We need to evaluate the integral $\int \frac{dx}{\sin^3(x)\cos^5(x)}$.

1. Problem Description

We need to evaluate the integral $\int \frac{dx}{\sin^3(x)\cos^5(x)}$.

2. Solution Steps

First, we can rewrite the integral as

$$$$

\int \frac{dx}{\sin^3(x)\cos^5(x)} = \int \frac{1}{\sin^3(x)\cos^5(x)} dx = \int \frac{\sin^2(x) + \cos^2(x)}{\sin^3(x)\cos^5(x)} dx

$$$$

We can split this into two integrals:

$$$$

\int \frac{\sin^2(x)}{\sin^3(x)\cos^5(x)} dx + \int \frac{\cos^2(x)}{\sin^3(x)\cos^5(x)} dx = \int \frac{1}{\sin(x)\cos^5(x)} dx + \int \frac{1}{\sin^3(x)\cos^3(x)} dx

$$$$

$$$$

= \int \frac{\cos(x)}{\sin(x)\cos^6(x)} dx + \int \frac{\sin(x)\cos(x)}{\sin^4(x)\cos^4(x)} dx

$$$$

Using the identity $\sin^2(x) + \cos^2(x) = 1$, we can write $1 = (\sin^2(x) + \cos^2(x))^n$ for any integer $n$. Let's multiply the integrand by $(\sin^2(x) + \cos^2(x))^3$.

$$$$

\int \frac{(\sin^2(x) + \cos^2(x))^3}{\sin^3(x)\cos^5(x)} dx = \int \frac{\sin^6(x) + 3\sin^4(x)\cos^2(x) + 3\sin^2(x)\cos^4(x) + \cos^6(x)}{\sin^3(x)\cos^5(x)} dx

$$$$

$$$$

= \int \left(\frac{\sin^3(x)}{\cos^5(x)} + \frac{3\sin(x)}{\cos^3(x)} + \frac{3}{\sin(x)\cos(x)} + \frac{\cos(x)}{\sin^3(x)}\right) dx

$$$$

$$$$

= \int \frac{\sin^3(x)}{\cos^5(x)} dx + \int \frac{3\sin(x)}{\cos^3(x)} dx + \int \frac{3}{\sin(x)\cos(x)} dx + \int \frac{\cos(x)}{\sin^3(x)} dx

$$$$

$$$$

= \int \tan^3(x)\sec^2(x)\sec(x) dx + 3\int \frac{\sin(x)}{\cos^3(x)} dx + 3\int \frac{1}{\sin(x)\cos(x)} dx + \int \frac{\cos(x)}{\sin^3(x)} dx

$$$$

Note that $\int \frac{1}{\sin(x)\cos(x)}dx = \int \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} dx = \int \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} dx = \int \tan(x) dx + \int \cot(x) dx = -\ln|\cos(x)| + \ln|\sin(x)| + C = \ln|\tan(x)| + C$.

Let $I = \int \frac{dx}{\sin^3(x)\cos^5(x)}$. We can rewrite it as follows:

$$$$

I = \int \frac{dx}{\sin^3(x)\cos^5(x)} = \int \frac{dx}{\frac{\sin^3(x)}{\cos^3(x)}\cos^8(x)} = \int \frac{\sec^8(x)}{\tan^3(x)}dx

$$$$

Let $u = \tan(x)$, then $du = \sec^2(x)dx$. We can rewrite $\sec^8(x)$ as $\sec^6(x)\sec^2(x) = (1+\tan^2(x))^3 \sec^2(x) = (1+u^2)^3 \sec^2(x)$.

$$$$

I = \int \frac{(1+u^2)^3}{u^3} du = \int \frac{1+3u^2+3u^4+u^6}{u^3} du = \int \left(\frac{1}{u^3} + \frac{3}{u} + 3u + u^3\right)du

$$$$

$$$$

I = \int u^{-3} + \frac{3}{u} + 3u + u^3 du = -\frac{1}{2}u^{-2} + 3\ln|u| + \frac{3}{2}u^2 + \frac{1}{4}u^4 + C

$$$$

$$$$

I = -\frac{1}{2\tan^2(x)} + 3\ln|\tan(x)| + \frac{3}{2}\tan^2(x) + \frac{1}{4}\tan^4(x) + C

$$$$

$$$$

I = -\frac{\cot^2(x)}{2} + 3\ln|\tan(x)| + \frac{3}{2}\tan^2(x) + \frac{\tan^4(x)}{4} + C

$$$$

3. Final Answer

$-\frac{1}{2\tan^2(x)} + 3\ln|\tan(x)| + \frac{3}{2}\tan^2(x) + \frac{1}{4}\tan^4(x) + C$